Answer :
Final answer:
The wavelengths of the first three lines in the Lyman series, for transitions to the n = 1 orbit with initial states n_i = 2, 3, and 4, are 121.6 nm, 102.6 nm, and 97.3 nm respectively. These calculations use the Rydberg equation for hydrogen's spectral lines.
Explanation:
To calculate the wavelengths of the first three lines in the Lyman series, we use the Rydberg equation:
\[\frac{1}{\lambda} = R \left( \frac{1}{1^2} - \frac{1}{n_i^2} \right)\]
Where \(\lambda\) is the wavelength, \(R\) is the Rydberg constant (1.097 x 10^7 m^-1), \(n_i\) is the initial energy level, and we are considering transitions to the n = 1 orbit (ground state).
For the first line (n_i = 2):
\[\frac{1}{\lambda} = 1.097 \times 10^7 \left(1 - \frac{1}{2^2}\right)\]
\[\lambda = 121.6 \text{ nm}\]
For the second line (n_i = 3):
\[\frac{1}{\lambda} = 1.097 \times 10^7 \left(1 - \frac{1}{3^2}\right)\]
\[\lambda = 102.6 \text{ nm}\]
For the third line (n_i = 4):
\[\frac{1}{\lambda} = 1.097 \times 10^7 \left(1 - \frac{1}{4^2}\right)\]
\[\lambda = 97.3 \text{ nm}\]
Thus, the wavelengths of the first three lines in the Lyman series in decreasing order are: 121.6 nm, 102.6 nm, and 97.3 nm. Therefore, the correct answer is b) 121.6nm, 102.6nm, 97.3nm.
